question_answer

The charge on the \[\text{4 }\!\!\mu\!\!\text{ F}\]capacitor in the steady state is

A)  \[\frac{20}{3}\,\mu C\]                

B) \[\frac{10}{3}\,\mu C\]

C)  \[\frac{5}{6}\,\mu C\]                  

D) \[10\,\mu C\]

Correct Answer: A

Solution :

 In the steady state, the capacitors are fully charged and act as an open circuit, so the equivalent circuit in steady state would be as shown in figure. Steady state current,\[I=\frac{10}{5+5}=1A\] So, potential drop across AB, \[V=5\text{ }V\] Sum of potential difference across \[2\mu F\]and \[4\mu F\] capacitors is 5 V. As capacitors are in series, charges on them would be the same. Let us say it is q. From KVL \[\frac{q}{2}+\frac{f}{4}=5\] \[\Rightarrow \]                               \[q=\frac{20}{3}\mu C\]